This problem here would be a little tricky. Let us take into account first the variables presented which are the following: a collection of triangular and square tiles, 25 tiles, and 84 edges. Triangles and squares are 2D in shape so they give us a variable of 3 and 4 to work on those edges. Let us say that we represent square tiles with x and triangular tiles with y. There would be two equations which look like these:
x + y = 25 and 4x + 3y = 84
The first one would refer to the number of tiles and the second one to number of edges.
We will be using the first equation to the second equation and solve for one. So if we will be looking for y for instance, then x in the second equation would be substituted with x = 25 - y which would look like this:
4 (25 - y) + 3y = 84
Solve.
100 - 4y + 3y = 84
-4y +3y = 84 - 100
-y = -16
-y/-1 = -16/-1
y = 16
Then:
x = 25 -y
x = 25 - 16
x = 9
So the answer is that there are 9 square tiles and 16 triangular tiles.
Let the three numbers be x, y, and z.
If the sum of the three numbers is 3, then x+y+z=3
If subtracting the second number from the sum of the first and third numbers gives 9, then x+z-y=9
If subtracting the third number from the sum of the first and second numbers gives -5, then x+y-z=-5
This forms the system of equations:
[1] x+y+z=3
[2] x-y+z=9
[3] x+y-z=-5
First, to find y, let's take do [1]-[2]:
x+y+z=3
-x+y-z=-9
2y=-6
y=-3
Then, to find z, let's do [1]-[3]:
x+y+z=3
-x+-y+z=5
2z=8
z=4
Now that you have y and z, plug them into [1] to find x:
x+y+z=3
x-3+4=3
x=2
So the three numbers are 2,-3, and 4.
Answer:
Tn=a+(n-1)d
a=a1
Tn=a1+(n+1)d
I would appreciate if my answer is chosen as a brainliest answer
Answer:
Incorrect
Step-by-step explanation:
This interpretation is incorrect because it states that 98% of the data is with in the confidence interval.
or 98% of the laptop have screen size between 19.2 and 21.4 inches
However, the interpretation would have been correct if it would have stated as - Value of the population mean i.e mean size of the laptop screen lies within the confidence interval.
